Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> y

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
A2(x, y) -> B2(0, c1(y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))
C1(b2(y, c1(x))) -> A2(0, 0)
C1(b2(y, c1(x))) -> B2(a2(0, 0), y)
A2(x, y) -> C1(y)
A2(x, y) -> B2(x, b2(0, c1(y)))

The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
A2(x, y) -> B2(0, c1(y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))
C1(b2(y, c1(x))) -> A2(0, 0)
C1(b2(y, c1(x))) -> B2(a2(0, 0), y)
A2(x, y) -> C1(y)
A2(x, y) -> B2(x, b2(0, c1(y)))

The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))
C1(b2(y, c1(x))) -> A2(0, 0)
A2(x, y) -> C1(y)

The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


C1(b2(y, c1(x))) -> A2(0, 0)
A2(x, y) -> C1(y)
The remaining pairs can at least be oriented weakly.

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(A2(x1, x2)) = 2 + x2   
POL(C1(x1)) = 1 + x1   
POL(a2(x1, x2)) = 2 + 2·x1 + x2   
POL(b2(x1, x2)) = x1 + x2   
POL(c1(x1)) = 2   

The following usable rules [14] were oriented:

b2(y, 0) -> y
a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

C1(b2(y, c1(x))) -> C1(b2(a2(0, 0), y))
C1(b2(y, c1(x))) -> C1(c1(b2(a2(0, 0), y)))

The TRS R consists of the following rules:

a2(x, y) -> b2(x, b2(0, c1(y)))
c1(b2(y, c1(x))) -> c1(c1(b2(a2(0, 0), y)))
b2(y, 0) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.